#pragma GCC optimize(2)
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <vector>

using namespace std;
using LL = long long;
/*
for each segment we can suppose that there are a roses and b lily
a + b = k               a = k - b               k = r - l + 1   
a * b = b * (k - b)  
if b = k/2   the value of a * b is max
so we can construct 1010101010101010....10 to satisfy this condition

*/

int n, m;

int main(){
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> m;
    for(int i = 1, x, y; i <= m; i ++) cin >> x >> y;

    for(int i = 1; i <= n; i ++){
        if(i % 2) cout << 1;
        else cout << 0;
    }



    return 0;
}